Question

A car accelerates uniformly from rest to 20 m/sec in 5.6 sec along a level stretch of road. Ignoring friction, determine the average power required to accelerate the car if (a) the weight of the car is 9,000 N, and (b) the weight of the car is 14,000 N.

Answer 1

Answer:

(a) $P=33000W$

(b) $P=51000W$

Explanation:

The average power is defined as the amount of work done during a time interval:

$P=\frac{W}{t}(1)$

According to work-energy theorem, the work done is equal to the change in kinetic energy. So, we have:

$W=\Delta K\\W=K_f-K_0\\W=\frac{mv_f^2}{2}-\frac{mv_0^2}{2}\\(2)$

Recall that the weight is given by:

$w=mg\\m=\frac{w}{g}(3)$

The car accelerates uniformly from rest ($v_0=0$). Replacing (3) in (2), we have:

$W=\frac{wv_f^2}{2g}$

(a) Finally, we replace this in (1):

$P=\frac{wv_f^2}{2gt}\\P=\frac{9000N(20\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(5.6s)}\\P=33000W$

(b)

$P=\frac{14000N(20\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(5.6s)}\\P=51000W$

Answer 2

(a) The average power required to accelerate the car of 9000 N is 32798.57 W.

(b)  The average power required to accelerate the car of 14,000 N is 51020.40 W.

Given data:

The initial velocity of car is, u = 0 m/s. (Since car was initially at rest)

The final velocity of car is, v = 20 m/s.

The time interval is, t = 5.6 s.

The given problem is based on the concept of average power. The average power is defined as the amount of work done during a time interval. Then,

P = W/t

Here, W is the work done and its value is obtained from the work - energy theorem as,

$W = \Delta KE\\\\W = \dfrac{1}{2}m(v^{2}-u^{2})$

Here, m is the mass.

(a)

For the weight of 9000 N, the mass of car is,

$w = mg\\\\9000 = m \times 9.8\\\\m =918.36 \;\rm kg$

So, the Work is obtained as,

$W =\dfrac{1}{2} \times 918.36 \times (20^{2}-0^{2})\\\\W =183672\;\rm J$

Then, the average power required to accelerate the car is,

P = W/t

P = 183672 / 5.6

P = 32798.57 W

Thus, we can conclude that the average power required to accelerate the car of 9000 N is 32798.57 W.

(b)

For the weight of 14,000 N, the mass of car is,

$w = mg\\\\14,000 = m \times 9.8\\\\m =1428.57 \;\rm kg$

So, the Work is obtained as,

$W =\dfrac{1}{2} \times 1428.57.36 \times (20^{2}-0^{2})\\\\W =285714.28\;\rm J$

Then, the average power required to accelerate the car is,

P = W/t

P = 285714.28 / 5.6

P = 51020.40 W

Thus, we can conclude that the average power required to accelerate the car of 14,000 N is 51020.40 W.

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